Cos alfa/2 - cos 3/2 alfa = 2 sin alfa sin gamma Ile wynosi gamma w zależności od alfy?.... Question from @Shovi

December 2018 1 19 Report

Cos alfa/2 - cos 3/2 alfa = 2 sin alfa sin gamma

Ile wynosi gamma w zależności od alfy?

Paawełek Wzór:

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Bez wzoru cos x - cos y widziałbym to tak:

$\hbox{Wzor ktory na pewno masz:} \\ \sin(2x) = 2\sin x \cos x \\ \hbox{Stad:} \\ \sin x =2\sin \dfrac{x}{2}\cos \dfrac{x}{2} \\ \hbox{Wzor ktory rowniez musisz miec:} \\ \cos(x+y)= \cos x \cos y - \sin x \sin y \\ \hbox{Stad:} \\ \cos \dfrac{3\alpha}{2}= \cos \left(\alpha+\dfrac{\alpha}{2}\right)=\cos \alpha \cos \dfrac{\alpha}{2}-\sin \alpha \sin \dfrac{\alpha}{2}$

więc masz równanie:
$\cos \dfrac{\alpha}{2} - \cos \alpha \cos \dfrac{\alpha}{2}+\sin \alpha \sin \dfrac{\alpha}{2}=4\sin \dfrac{\alpha}{2}\cos \dfrac{\alpha}{2}\sin \gamma \\ \\ \\ \cos \dfrac{\alpha}{2}(1-\cos \alpha) + \sin \alpha \sin \dfrac{\alpha}{2}=4\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2} \sin \gamma$

Wzór

$\cos (2x)=1-2\sin^2 x \\ \hbox{Stad:} \\ \cos x = 1-2\sin^2 \dfrac{x}{2} \\ \\ \cos x -1 = -2\sin^2 \dfrac{x}{2} \qquad /\cdot (-1) \\ \\ 1-\cos x = 2\sin^2 \dfrac{x}{2}$

Podkładam to i mam:

$\cos \dfrac{\alpha}{2} \cdot 2\sin^2 \dfrac{\alpha}{2}+ \sin \alpha \sin \dfrac{\alpha}{2}=4\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2} \sin \gamma \qquad /:\sin \dfrac{\alpha}{2} \\ \\ 2 \sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2}+\sin \alpha = 4\cos \dfrac{\alpha}{2} \sin \gamma \\ \\ 2 \sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2}+2\sin \dfrac{\alpha}{2} \cos\dfrac{\alpha}{2} = 4\cos \dfrac{\alpha}{2} \sin \gamma \qquad /:2\cos \dfrac{\alpha}{2} \\ \\$

$\sin \dfrac{\alpha}{2}+\sin \dfrac{\alpha}{2}=2\sin \gamma \\ \\ 2\sin \dfrac{\alpha}{2}=2\sin \gamma \qquad /:2 \\ \\ \sin \dfrac{\alpha}{2}=\sin \gamma \\ \\ \gamma = \dfrac{\alpha}{2}$